Lüroth's theorem

Remarks

Pick $u \in E \setminus K$ and write $u = f(t)/g(t)$ with $f, g$ coprime polynomials. Then $t$ is a root of $f(x) - u\, g(x)$ over $K(u)$, and one shows that $[K(t) : K(u)] = \max(\deg f, \deg g)$. The proof then takes the minimal polynomial of $t$ over $E$, chooses $u$ among its coefficients not in $K$, and compares degrees to get $E = K(u)$. Example: $E = K(t^2 + t^{-2})$ lies between $K$ and $K(t)$, and the theorem says it is already of the form $K(u)$; here $u = t^2 + t^{-2}$ works, and $[K(t) : E] = 4$. The analogue in two variables concerns fields $E \subseteq K(x, y)$ with $K(x, y)$ finite over $E$. It holds over algebraically closed fields of characteristic $0$ by Castelnuovo's theorem, and it fails in general in three or more variables.

Going deeper

For a student

The theorem says that a subfield of a purely transcendental extension of transcendence degree $1$ is purely transcendental. In geometry, it gives that a unirational curve is rational. The degree formula $[K(t) : K(u)] = \max(\deg f, \deg g)$ also computes the degree of a rational map from the line to itself, which is the degree of the field extension $K(t)/K(u)$.